# Electrical Engineering - Parallel Circuits - Discussion

Discussion Forum : Parallel Circuits - General Questions (Q.No. 22)
22.
Four resistors are connected in parallel. Fifteen mA flows through resistor R. If the second resistor is 2R, the third resistor 3R, and the fourth resistor 4R, the total current in the circuit is
60 mA
15 mA
135 mA
31.25 mA
Explanation:
No answer description is available. Let's discuss.
Discussion:
6 comments Page 1 of 1.

Someone please get me the explanation.

When resistance increases current decrease.

First way of solving
---------------------
I1=15--->R
I2=15/2---->2R
I3=15/3----->3R
I4=15/4------>4R
I=I1+I2+I3+I4=31.25

Second way of solving
---------------------
I1=(It*Rt)/R----(1)
It-->Total current
Rt-->Total resistance
It*Rt---->Voltage
Rt=1/(1/R+1/2R+1/3R+1/4R)=12R/25
From (1) equation
15=(It*12R/25)/R=It*12R/25*R=It*12/25
=>It=15*25/12=31.25
(1)

Twumasi samuel said:   1 decade ago
15*R = I*2R

I = 15/2 = 7.5.

= 15/3 = 5.

= 15/4 = 3.75.

TOTAL current is 15+7.5+5+3.75 = 31.25.
(2)

Kichu said:   8 years ago
Well said @Twumasi.

Maybe said:   7 years ago
Since voltage remains constant,
branch 1->V=IR,
branch 2->V=(I/2)2R,
So on.

Thus Itotal=15+(15/2)+(15/3)+(15/4)=31.25mA.

Maybe said:   7 years ago
Since voltage remains constant.
branch 1->V=IR.
branch 2->V=(I/2)2R.
So on
thus Itotal=15+(15/2)+(15/3)+(15/4)=31.25mA.
(1)